determine the constant of variation for the direct variation R varies directly with S is given 16, R is 80
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A function assigns the values. The number of people that were recruited in year 6 is 192.
<h3>What is a Function?</h3>A function assigns the value of each element of one set to the other specific element of another set.
Given that the number of people per year is represented by the function .
In the given function, t represents the number of years.
Now, the number of people that were recruited in year 6 can be calculated by putting the value of t as 6 in the given function,
f(6) = 3(2)⁶
= 3(64)
= 192
Hence, the number of people that were recruited in year 6 is 192.
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First, you have to find out the values of the length and width.
Let l = length and w = width.
Set up your equation for area: l * w = 80
Set up your equation for perimeter: 2l + 2w = 36
We can use substitution for this problem, so find out the value of l or w. In this case, I will use the first equation to find out the value of l.
(Divide w on both sides) l = 80 / w
Since I know that l = 80 / w, I can substitute it into the second equation.
2l + 2w = 36, and l = 80 / w
= 2 (80/w) + 2w = 36
Simplify the equation.
= 160/w + 2w = 36
Get rid of the w in the denominator by multiplying the entire equation by w.
= ( 160/w + 2w = 36 ) * w
= 160 + = 36w
= 2 ( w - 8 ) ( w - 10 )
The solutions for width (w) is 8 or 10. Then, you can use both 8 and 10 as your length and width, either or is fine. Multiply by 4 for each of those values and you get 8 x 4 = 32 and 10 x 4 = 40. The new perimeter would be 2(32) + 2(40) =
a. 144 cm
Hope this helped you! Let me know if there are any confusions :)